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Vikentia [17]
3 years ago
5

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Physics
2 answers:
Alekssandra [29.7K]3 years ago
5 0

A. Amperes is derived from coulomb and second.

B. Newton is derived from Kilogram, meter, and second.

Second and kilogram are base units.

Mkey [24]3 years ago
3 0

A and B are both derived from base units of metres kilograms seconds. I'd try A

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During aerobic activity, if your heart rate is lower than the lower limit, you are ___________________.
soldier1979 [14.2K]
Not pushing yourself hard enough is the answer since your heart rate doesn't even hit your lower minimum.
Pushing yourself to the limit is at your max heart rate.
Just at the right spot is at your max heart rate.
Pushing yourself too hard is above your max heart rate.
4 0
4 years ago
A Plane has a takeoff speed of 150 m/s and requires 1500m to reach that speed. Determine the acceleration of the plane and the t
mars1129 [50]

<u>Answer:</u>

The acceleration of the plane and the time required to reach this speed is  (a)= 7.5 m/sec^2 and time(t) = 20 seconds  

<u>Explanation: </u>

Given data Initial velocity (V_i) = 0  

Final velocity (V_f) = 150 m/second

Distance (d) = 1500 m

We have the formula,  $\mathrm{V}_{\mathrm{f}}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 \mathrm{ad}$

which gives 150^2 = 0+2a(1500)    

22500 = 3000 a  

acceleration (a) = 7.5 m/s^2

$\mathrm{V}_{\mathrm{f}}=\mathrm{V}_{\mathrm{i}}+\mathrm{at}$

150 = 7.5 t

t= 150/7.5 = 20

t = 20 seconds.  

5 0
3 years ago
Which wave property is directly related to energy
GrogVix [38]
The wave property is called frequency
5 0
3 years ago
Read 2 more answers
“Anchor” points on a temperature scale are known as fiducial points.<br><br> a)True<br> b)False
Svetlanka [38]
Yes. It r<span>efers to any of the temperatures assigned to a number of reproducible equilibrium states on the International Practical Temperature Scale</span><span>

In short, Your Answer would be "True"

Hope this helps!</span>
8 0
3 years ago
Read 2 more answers
Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. one force is 9.0 n, acting due east, an
Zolol [24]
The first thing you should do for this case is to find the horizontal and vertical components of the forces acting on the body.
 We have then:
 Horizontal = 9-9.2cos (58) = 4.124742769 N.
 Vertical = 9.2sin (58) = 7.802042485 N
 Then, the resulting net force is:
 F = √ ((4.124742769) ^ 2 + (7.802042485) ^ 2) = 8.825268826 N
 Then by definition:
 F = m * a
 Clearing the acceleration:
 a = F / m
 a = (8.825268826) / (3.0) = 2.941756275 m / s ^ 2
 answer:
 The magnitude of the body's acceleration is
 2.941756275 m / s ^ 2
6 0
3 years ago
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